Ferkans — Interactive Telecom Tutor

The Active-Passive Crossover Distance

Active RIS has advantages (breaking the $d^2 d^2$ ceiling) and disadvantages (amplifier noise, power consumption, hardware complexity). At what distance does active RIS become the better choice? The answer is a "crossover distance" $d^\star$ that depends on $N, g_{\max}, \sigma^2_{\text{RIS}}$ , and the direct path. Below this distance, passive RIS suffices; above it, active is needed. This section derives the crossover formula and explains its scaling.

Theorem: Active-Passive Crossover Distance

Let $d_0 = d_1 + d_2$ be the BS-UE distance. Under symmetric geometry ( $d_1 = d_2 = d_0/2$ ), equal-amplitude per-hop channels $\alpha^2 = \beta^2 = \lambda^{2}/(4\pi d/2)^2$ , $N$ RIS elements with max amplifier gain $g_{\max}$ and amplifier noise variance $\sigma^2_{\text{RIS}}$ , the crossover distance $d^\star$ where active SNR equals passive SNR satisfies

$d^{\star\,2} = \frac{g_{\max}^2 - 1}{\sigma^2_{\text{RIS}} / \sigma^2} \cdot \frac{4\pi^2}{\lambda^{2}}.$

For $d < d^\star$ : passive wins. For $d > d^\star$ : active wins. The crossover scales linearly with $g_{\max}$ (amplifier gain) and inversely with the amplifier noise figure (noise variance of the active component).

Passive RIS SNR falls as $d_1^2 d_2^2$ (product path loss). Active RIS SNR at high gain is independent of $d_2$ (the RIS-UE distance) because the amplified noise scales with $d_2^2$ , canceling the $d_2^2$ in the signal. Below some crossover, passive's $N^2$ gain beats active's flat noise floor; above it, active wins.

Proof

Passive SNR

Passive coherent SNR: $\text{SNR}_{P} = P_t N^2 \alpha^2 \beta^2 / \sigma^2$ . Substituting $\alpha^2 = \beta^2 = \lambda^{2}/(4\pi d/2)^2$ : $\text{SNR}_{P} = P_t N^2 \lambda^{4} / (16 \pi^2 (d/2)^4 \sigma^2) = P_t N^2 \lambda^{4} / (\pi^2 d^4 \sigma^2)$ .

Active SNR at high gain

From Theorem 9.2: $\text{SNR}_{A} \approx P_t N \alpha^2 / \sigma^2_{\text{RIS}}$ (large $g_{\max}$ ). Substituting: $\text{SNR}_{A} \approx P_t N \lambda^{2} / (\pi^2 d^2 \sigma^2_{\text{RIS}})$ .

Crossover

Setting $\text{SNR}_{P} = \text{SNR}_{A}$ : $N \lambda^{2} / d^2 = \sigma^2_{\text{RIS}} / \sigma^2$ (dropping common factors). Hence $d^{\star\,2} = N \lambda^{2} \sigma^2 / \sigma^2_{\text{RIS}}$ , or $d^\star = \lambda \sqrt{N \cdot (\sigma^2/\sigma^2_{\text{RIS}})}$ . Higher $N$ and lower RIS noise push the crossover to larger distance (passive wins over a longer range). $\blacksquare$

,

Key Takeaway

Active RIS wins for long-distance links; passive wins for short. The crossover is at roughly $d^\star \sim \lambda\sqrt{N/\text{NF}}$ where NF is the amplifier noise figure. At 28 GHz with $N = 256$ and NF = 5 dB: $d^\star \sim 0.01 \cdot \sqrt{256/3.16} \approx 0.09\,\text{m}$ . For practical mmWave deployments at $10$ - $100\,\text{m}$ range, active is often needed. At lower frequencies (sub-6 GHz) with larger $\lambda$ and lower path loss, passive goes further.

Active-Passive SNR Crossover

For varying BS-UE distance $d_0$ , plot the SNR of active and passive RIS alongside the direct link. The crossover point depends on $N$ , amplifier gain, noise figure. Change parameters to see how the crossover shifts.

Parameters

RIS elements

N

256

Carrier frequency (GHz)28

Max amplifier gain (dB)20

Amplifier noise figure (dB)5

Max distance swept (m)100

Example: Crossover at mmWave: Active RIS Wins Beyond 30 m

mmWave scenario: $f_c = 28\text{ GHz}, \lambda = 1.07\text{ cm}$ . $N = 256$ , $g_{\max} = 20\text{ dB}$ , amplifier noise figure $5\text{ dB}$ (so $\sigma^2_{\text{RIS}} / \sigma^2 = 3.16$ ). Compute the crossover distance.

Solution

Use formula

$d^{\star\,2} = N \lambda^{2} \cdot (\sigma^2/\sigma^2_{\text{RIS}}) = 256 \cdot (0.0107)^2 \cdot (1/3.16)$ $= 256 \cdot 0.000114 \cdot 0.316 \approx 0.00923$ . So $d^\star \approx 0.096\text{ m}$ (!) — very short.

Sanity check

$\sim 10$ cm is too short for practical deployments. This says that at mmWave, passive RIS is essentially never competitive — active is needed. But the formula uses idealizations; the practical crossover incorporates BS gain, antenna patterns, and is usually $1$ - $30\,\text{m}$ depending on geometry.

Practical value

Realistic mmWave deployment with $N_t = 16$ BS antennas (aperture gain $16$ ), moderate BS-RIS distance (few meters), $N = 256$ : active RIS dominates at all operational distances. Passive is a research curiosity at these frequencies. Sub-6 GHz is different: passive can go tens of meters before active becomes needed.

A Three-Regime Operating Picture

Active-RIS deployment falls into three regimes:

Short distance ( $d < d^\star$ ): passive RIS is sufficient and cheaper. Active amplifier noise erodes gains.
Crossover region ( $d \approx d^\star$ ): the choice depends on secondary factors: power availability (active needs DC), deployment cost, update rate. Either can work.
Long distance ( $d > d^\star$ ): passive's product-path-loss ceiling means it simply can't close the link. Active is the only RIS option.

The "correct" architecture choice depends on the deployment's typical UE distance distribution. Hybrid deployments (active +passive) serving different user populations are a natural extension.

Common Mistake: Don't Forget Active RIS Consumes DC Power

Mistake:

"Active RIS outperforms passive at mmWave, so let's use it everywhere."

Correction:

Active RIS consumes $\sim 1$ - $5$ W of DC per panel (Section 9.1's engineering note). Passive consumes microwatts. Over the lifetime of a deployment, the DC cost is substantial — and active RIS requires a wired power connection, limiting where it can be installed. Weigh the SNR gain against the deployment overhead. In many scenarios, a larger passive RIS at lower $B$ (discrete phases, $N = 1024$ ) is cheaper and more robust than a smaller active RIS at $N = 256$ with $g_{\max} = 20$ dB.

When Does Active Beat Passive?